# 高中數學/函數與三角/三角恆等變換

## 基礎知識

### 二倍角公式

• ${\displaystyle \sin 2\theta =2\sin \theta \cos \theta }$
• ${\displaystyle \cos 2\theta =\cos ^{2}\theta -\sin ^{2}\theta =2\cos ^{2}\theta -1=1-2\sin ^{2}\theta }$
• ${\displaystyle \tan 2\theta ={\frac {2\tan \theta }{1-\tan ^{2}\theta }}}$

A.${\displaystyle \sin 2-\cos 2}$；B.${\displaystyle \cos 2-\sin 2}$；C.${\displaystyle \cos 2}$；D.${\displaystyle -\cos 2}$

(1) ${\displaystyle (\cos {\frac {\pi }{12}}-\sin {\frac {\pi }{12}})(\cos {\frac {\pi }{12}}+\sin {\frac {\pi }{12}})}$
(2) ${\displaystyle \sin ^{4}{\frac {\pi }{12}}-\cos ^{2}{\frac {\pi }{12}}}$
(3) ${\displaystyle {\sqrt {2-\sin ^{2}2+\cos 4}}}$
(4) ${\displaystyle {\frac {\sqrt {3}}{\cos 10^{\circ }}}-{\frac {1}{\sin 170^{\circ }}}}$
(5) ${\displaystyle {\frac {1-\tan ^{2}75^{\circ }}{\tan 75^{\circ }}}}$
(6) ${\displaystyle {\frac {\tan 14^{\circ }}{1-\tan ^{2}14^{\circ }}}\cdot \cos 28^{\circ }}$
(7) ${\displaystyle \sin 50^{\circ }(1+{\sqrt {3}}\tan 10^{\circ })}$
(8) ${\displaystyle 4\cos 50^{\circ }-\tan 40^{\circ }}$
(9) ${\displaystyle \tan {\frac {\pi }{8}}-\cot {\frac {\pi }{8}}}$

A.${\displaystyle 2\sin 15^{\circ }\cos 15^{\circ }}$；B.${\displaystyle 1-2\sin ^{2}15^{\circ }}$；C.${\displaystyle \sin ^{2}15^{\circ }+\cos ^{2}15^{\circ }}$；D.${\displaystyle {\frac {3\tan 15^{\circ }}{1-\tan ^{2}15^{\circ }}}}$

（提示：此題也可以使用柯西不等式求解。）

A.${\displaystyle {\frac {n}{1+m}}}$；B.${\displaystyle {\frac {m}{1+n}}}$；C.${\displaystyle {\frac {1-n}{m}}}$；D.${\displaystyle {\frac {1-m}{n}}}$

(1) 求${\displaystyle \sin(2A-{\frac {\pi }{2}})}$的值。
(2) 求${\displaystyle \cos B}$的值。

(1) 求${\displaystyle \tan 2x}$的值。
(2) 求${\displaystyle \cos(2x-{\frac {\pi }{4}})}$的值。

(1) 求${\displaystyle \tan A}$的值。
(2) 求${\displaystyle \cos(A+B)}$的值。

${\displaystyle {\begin{array}{l}{\frac {\sin x-2\cos x}{\sin x+\cos x}}=2\\\Rightarrow \quad \sin x-2\cos x=2\sin x+2\cos x\\\Rightarrow \quad \sin x=-4\cos x\end{array}}}$

${\displaystyle \sin 2x=1-2\sin ^{2}x=1-2\cdot {\frac {1}{17}}=1-{\frac {2}{17}}={\frac {15}{17}}}$

(1) 求${\displaystyle f(\pi )}$的值。
(2) 已知${\displaystyle t\in (-{\frac {\pi }{2}},0),f(t+{\frac {2\pi }{3}})={\frac {6}{5}}}$，求${\displaystyle f(2t)}$的值。

(1) 求${\displaystyle \cos(2A+{\frac {\pi }{4}})}$的值。
(2) 已知${\displaystyle A\in (0,{\frac {\pi }{2}}),B\in (-{\frac {\pi }{2}},0),\sin(B+{\frac {\pi }{4}})={\frac {\sqrt {10}}{10}}}$，求${\displaystyle A-B}$的值。

### 半角公式與冪的升降

• ${\displaystyle \sin \theta =2\sin({\frac {\theta }{2}})\cos({\frac {\theta }{2}})}$
• ${\displaystyle \cos \theta =2\cos ^{2}({\frac {\theta }{2}})-1=1-2\sin ^{2}({\frac {\theta }{2}})=\cos ^{2}({\frac {\theta }{2}})-\sin ^{2}({\frac {\theta }{2}})}$
• ${\displaystyle \tan \theta ={\frac {2\tan({\frac {\theta }{2}})}{1-\tan ^{2}({\frac {\theta }{2}})}}}$

• ${\displaystyle \sin ^{2}\theta ={\frac {1-\cos 2\theta }{2}}}$
• ${\displaystyle \cos ^{2}\theta ={\frac {1+\cos 2\theta }{2}}}$
• ${\displaystyle \tan ^{2}\theta ={\frac {1-\cos 2\theta }{1+\cos 2\theta }}}$

(1) 求${\displaystyle \sin x}$的值。
(2) 求${\displaystyle \cos ^{2}({\frac {\pi }{4}}-{\frac {x}{2}})+\sin(3\pi +{\frac {x}{2}})\sin({\frac {3\pi }{2}}-{\frac {x}{2}})}$的值。

(1) 求${\displaystyle \cos {\frac {A-B}{2}}}$的值。
(2) 求${\displaystyle \tan {\frac {A-B}{2}}}$的值。

### 輔助角公式

${\displaystyle {\begin{array}{l}a\sin x+b\cos x\\={\sqrt {a^{2}+b^{2}}}({\frac {a}{\sqrt {a^{2}+b^{2}}}}\sin x+{\frac {b}{\sqrt {a^{2}+b^{2}}}}\cos x)\\={\sqrt {a^{2}+b^{2}}}(\cos \phi \sin x+\sin \phi \cos x)\\={\sqrt {a^{2}+b^{2}}}\sin(x+\phi )\end{array}}}$

A.${\displaystyle y=\sin 2x+\cos 2x}$；B.${\displaystyle y=\sin 2x\cos 2x}$；C.${\displaystyle y=\cos(4x+{\frac {\pi }{2}})}$；D.${\displaystyle y=\sin ^{2}2x-\cos ^{2}2x}$

${\displaystyle {\begin{array}{l}2\cos ^{2}x+\sin 2x=2\cdot {\frac {1+\cos 2x}{2}}+\sin 2x\\=1+\cos 2x+\sin 2x\\=1+2(\sin 2x\cdot {\frac {1}{2}}+\cos 2x\cdot {\frac {1}{2}})\\=1+2\sin(2x+{\frac {\pi }{4}})\end{array}}}$

${\displaystyle {\begin{array}{l}y={\frac {1}{2}}\sin 2x+{\frac {1-\cos 2x}{2}}\\={\frac {1}{2}}\sin 2x-\cos 2x{\frac {1}{2}}+{\frac {1}{2}}=(\sin 2x\cdot {\frac {1}{2}}-\cos 2x\cdot {\frac {1}{2}})+{\frac {1}{2}}=(\sin 2x\cos {\frac {\pi }{4}}-\cos 2x\sin {\frac {\pi }{4}})+{\frac {1}{2}}=\sin(2x-{\frac {\pi }{4}})+{\frac {1}{2}}\end{array}}}$

${\displaystyle {\begin{array}{l}y=(\sin x\cos {\frac {\pi }{6}}-\cos x\sin {\frac {\pi }{6}})\cos x\\=(\sin x\cdot {\frac {\sqrt {3}}{2}}-\cos x\cdot {\frac {1}{2}})\cos x\\={\frac {\sqrt {3}}{2}}\sin x\cos x-{\frac {1}{2}}\cos ^{2}x\\={\frac {\sqrt {3}}{4}}\cdot 2\sin x\cos x-{\frac {1}{2}}\cdot {\frac {1+\cos 2x}{2}}\\={\frac {1}{2}}({\frac {\sqrt {3}}{2}}\sin 2x-{\frac {1}{2}}\cos 2x)+{\frac {1}{4}}\\={\frac {1}{2}}(\cos {\frac {\pi }{6}}\sin 2x-\sin {\frac {\pi }{6}}\cos 2x)+{\frac {1}{4}}\\={\frac {1}{2}}\sin(2x-{\frac {\pi }{6}})+{\frac {1}{4}}\\\leq {\frac {1}{2}}\cdot 1+{\frac {1}{4}}={\frac {3}{4}}\end{array}}}$

### 三角恆等變換常用公式匯總

${\displaystyle \cos(\alpha \pm \beta )=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta }$
${\displaystyle \tan(\alpha \pm \beta )={\frac {\tan \alpha \pm \tan \beta }{1\mp \tan \alpha \tan \beta }}}$

${\displaystyle \cos 2\theta =\cos ^{2}\theta -\sin ^{2}\theta =2\cos ^{2}\theta -1=1-2\sin ^{2}\theta }$
${\displaystyle \tan 2\theta ={\frac {2\tan \theta }{1-\tan ^{2}\theta }}}$

${\displaystyle \cos \theta =2\cos ^{2}({\frac {\theta }{2}})-1=1-2\sin ^{2}({\frac {\theta }{2}})=\cos ^{2}({\frac {\theta }{2}})-\sin ^{2}({\frac {\theta }{2}})}$
${\displaystyle \tan \theta ={\frac {2\tan({\frac {\theta }{2}})}{1-\tan ^{2}({\frac {\theta }{2}})}}}$

${\displaystyle \cos ^{2}\theta ={\frac {1+\cos 2\theta }{2}}}$
${\displaystyle \tan ^{2}\theta ={\frac {1-\cos 2\theta }{1+\cos 2\theta }}}$

(1) 求${\displaystyle \cos A}$的值。
(2) 求${\displaystyle \cos B}$的值。

(1) 求${\displaystyle \cos x}$${\displaystyle \tan(x+\pi )}$的值。
(2) 求${\displaystyle \sin(x+{\frac {\pi }{4}})}$${\displaystyle \cos(x-{\frac {\pi }{3}})}$的值。

(1) ${\displaystyle \sin(\alpha +\beta )\cdot \sin(\alpha -\beta )=\sin ^{2}\alpha -\sin ^{2}\beta }$
(2) ${\displaystyle \cos(\alpha +\beta )\cdot \cos(\alpha -\beta )=\cos ^{2}\alpha -\sin ^{2}\beta }$

## 常用結論與常見模型

### 弦化切的技巧與數字「1」的轉換

(1) 求${\displaystyle \tan(x+{\frac {\pi }{6}})}$的值。
(2) 求${\displaystyle \cos(2x+{\frac {7\pi }{12}})}$的值。

(1) ${\displaystyle {\sqrt {1+\sin 2}}}$
(2) ${\displaystyle {\frac {\cos 20^{\circ }}{\cos 35^{\circ }{\sqrt {1-\sin 20^{\circ }}}}}}$

(1) 求${\displaystyle \tan x}$的值。
(2) 求${\displaystyle {\frac {\cos 2x}{\sin 2x+1}}}$的值。

### 角的拼拆

(1) 求函數${\displaystyle f(x)}$的最大值以及取得最大值時相應的x的集合。
(2) 若${\displaystyle t\in ({\frac {\pi }{4}},{\frac {\pi }{2}}),f(t)={\frac {4}{5}}}$，求${\displaystyle \cos 2t}$的值。

(1)
${\displaystyle {\begin{array}{l}f(x)=\sin(2x-{\frac {\pi }{6}})+2\cos ^{2}x-1\\=\sin(2x-{\frac {\pi }{6}})+(2\cos ^{2}x-1)\\=(\sin 2x\cos {\frac {\pi }{6}}-\cos 2x\sin {\frac {\pi }{6}})+\cos 2x\\=(\sin 2x\cdot {\frac {\sqrt {3}}{2}}-\cos 2x\cdot {\frac {1}{2}})+\cos 2x\\=\sin 2x\cdot {\frac {\sqrt {3}}{2}}+\cos 2x\cdot {\frac {1}{2}}\\=\sin 2x\cos {\frac {\pi }{6}}+\cos 2x\sin {\frac {\pi }{6}}=\sin(2x+{\frac {\pi }{6}})\leq 1\end{array}}}$

(2) 由前一問所得的化簡結果和t的已知範圍可知：
${\displaystyle f(t)=\sin(2t+{\frac {\pi }{6}})\quad (t\in ({\frac {\pi }{4}},{\frac {\pi }{2}})}$

${\displaystyle {\begin{array}{l}\sin(2t+{\frac {\pi }{6}})={\frac {4}{5}}\\\Rightarrow \quad \sin 2t\cos {\frac {\pi }{6}}+\cos 2t\sin {\frac {\pi }{6}}={\frac {4}{5}}\\\Rightarrow \quad \sin 2t\cdot {\frac {\sqrt {3}}{2}}+\cos 2t\cdot {\frac {1}{2}}={\frac {4}{5}}\\\Rightarrow \quad {\sqrt {3}}\sin 2t+\cos 2t={\frac {8}{5}}\end{array}}}$

${\displaystyle {\begin{array}{l}{\sqrt {3}}{\sqrt {1-\cos ^{2}2t}}+\cos 2t={\frac {8}{5}}\\{\sqrt {3}}{\sqrt {1-\cos ^{2}2t}}={\frac {8}{5}}-\cos 2t\\3({\sqrt {1-\cos ^{2}2t}})^{2}=({\frac {8}{5}}-\cos 2t)^{2}\\3(1-\cos ^{2}2t)=({\frac {8}{5}})^{2}-{\frac {16}{5}}\cos 2t+\cos ^{2}2t\\4\cos ^{2}2t-{\frac {16}{5}}\cos 2t-{\frac {11}{25}}=0\end{array}}}$

(2) 根據前面的討論和已知條件，容易得到：
${\displaystyle \sin(2t+{\frac {\pi }{6}})={\frac {4}{5}}}$

${\displaystyle {\begin{array}{l}\cos(2t+{\frac {\pi }{6}})=-{\sqrt {1-\sin ^{2}(2t+{\frac {\pi }{6}})}}=-{\sqrt {1-({\frac {4}{5}})^{2}}}=-{\frac {3}{5}}\\\Rightarrow \quad \cos 2t=\cos((2t+{\frac {\pi }{6}})-{\frac {\pi }{6}})=\cos(2t+{\frac {\pi }{6}})\cos {\frac {\pi }{6}}+\sin(2t+{\frac {\pi }{6}})\sin {\frac {\pi }{6}}\\={\frac {4}{5}}\cdot {\frac {1}{2}}+(-{\frac {3}{5}})\cdot {\frac {\sqrt {3}}{2}}={\frac {4}{10}}-{\frac {3{\sqrt {3}}}{10}}={\frac {4-3{\sqrt {3}}}{10}}\end{array}}}$

(1) 求函數${\displaystyle f(x)}$的最大值及取得最大值時x的集合。
(2) 若${\displaystyle t\in ({\frac {\pi }{4}},{\frac {\pi }{2}}),f(t)={\frac {2}{5}}}$，求${\displaystyle \sin 2t}$的值。

(1)
${\displaystyle {\begin{array}{l}f(x)=\cos 2x\cos {\frac {\pi }{3}}-\sin 2x\sin {\frac {\pi }{3}}+2{\frac {1-\cos 2x}{2}}\\=-({\frac {\sqrt {3}}{2}}\sin 2x+{\frac {1}{2}}\cos 2x)+1\\=-\sin(2x+{\frac {\pi }{6}})+1\leq -(-1)+1=2\end{array}}}$

(2) 因為${\displaystyle f(t)={\frac {3}{5}}}$，所以${\displaystyle \sin(2t+{\frac {\pi }{6}})={\frac {3}{5}}}$

${\displaystyle {\begin{array}{l}\sin 2t=\sin((2t+{\frac {\pi }{6}})-{\frac {\pi }{6}})=\sin(2t+{\frac {\pi }{6}})\cos {\frac {\pi }{6}}-\cos(2t+{\frac {\pi }{6}})\sin {\frac {\pi }{6}}\\={\frac {3}{5}}\cdot {\frac {\sqrt {3}}{2}}-(-{\frac {4}{5}})\cdot {\frac {1}{2}}={\frac {3{\sqrt {3}}}{10}}+{\frac {4}{10}}={\frac {4+3{\sqrt {3}}}{10}}\end{array}}}$

### 三角函數的單調性與圖象性質

A.存在${\displaystyle x_{1},x_{2}}$，當${\displaystyle x_{1}-x_{2}=\pi }$時，有${\displaystyle f(x_{1})=f(x_{2})}$成立。
B.${\displaystyle f(x)}$在區間${\displaystyle [{\frac {-\pi }{6}},{\frac {\pi }{3}}]}$上單調遞增。
C.函數${\displaystyle f(x)}$的圖象關於點${\displaystyle ({\frac {\pi }{12}},0)}$對稱。
D.函數${\displaystyle f(x)}$的圖象關於直線${\displaystyle x={\frac {5\pi }{12}}}$對稱。

(1) 求${\displaystyle f(x)}$的最小正周期。
(2) 分析${\displaystyle f(x)}$在區間${\displaystyle [0,\pi ]}$上的單調性。

(1) 求函數${\displaystyle f(x)}$的最小正周期和單調遞增區間。
(2) 若${\displaystyle x_{1},x_{2}}$是函數${\displaystyle y=f(x)-{\frac {1}{2}}}$的2個零點，求${\displaystyle |x_{1}-x_{2}|}$的最小值。

### 三角換元法

${\displaystyle {\begin{array}{l}f(x)=x+{\sqrt {1-x^{2}}}\\=\cos \theta +{\sqrt {1-\sin ^{2}\theta }}\\=\cos \theta +{\sqrt {\cos ^{2}\theta }}\\=\cos \theta +\cos \theta \\=2\cos \theta \leq 2\cdot 1=2\end{array}}}$

### 涉及三角形的問題

A.等邊三角形；B.等腰三角形；C.等腰直角三角形；D.直角三角形

A.等腰三角形；B.直角三角形；C.等腰直角三角形；D.等腰三角形或直角三角形

## 補充習題

• 下列各式中，結果為1的是(    )。
A.${\displaystyle {\frac {22.5^{\circ }}{1-\tan ^{2}22.5^{\circ }}}}$；B.${\displaystyle \tan 15^{\circ }\cos ^{2}15^{\circ }}$；C.${\displaystyle {\frac {\sqrt {3}}{3}}\cos ^{2}{\frac {\pi }{12}}-{\frac {\sqrt {3}}{3}}\sin ^{2}{\frac {\pi }{12}}}$；D.${\displaystyle {\sqrt {\frac {1-\cos 60^{\circ }}{2}}}}$
• 下列各式與${\displaystyle \tan x}$相等的是(    )。
A.${\displaystyle {\sqrt {\frac {1-\cos 2x}{1+\cos 2x}}}}$；B.${\displaystyle {\frac {\sin x}{1+\cos x}}}$
C.${\displaystyle {\frac {1-\cos 2x}{\sin 2x}}}$；D.${\displaystyle {\sqrt {\frac {1+\cos(\pi +2x)}{2}}}\cdot {\frac {1}{\cos x}}\quad (x\in (0,\pi ))}$
• 計算或化簡下列各式：
(1) ${\displaystyle \tan 9^{\circ }+\cot 117^{\circ }-\tan 243^{\circ }-\cot 351^{\circ }}$
(2) ${\displaystyle \tan 15^{\circ }-\cot 15^{\circ }}$
(3) ${\displaystyle {\frac {\sqrt {3}}{\sin 20^{\circ }}}-{\frac {1}{\cos 20^{\circ }}}}$
• 已知${\displaystyle \tan x={\frac {1}{3}}}$，求${\displaystyle {\frac {\sin x+2\cos x}{5\cos x-\sin x}}}$的值。
• 已知A是函數${\displaystyle f(x)=2\sin(2018x+{\frac {\pi }{4}})+\cos(2018x-{\frac {\pi }{4}})}$的最大值。若存在實數${\displaystyle x_{1},x_{2}}$，使得對於任意實數x總有${\displaystyle f(x_{1})\leq f(x)\leq f(x_{2})}$成立，求${\displaystyle A\cdot |x_{1}-x_{2}|}$的最小值。
• 已知下列3個式子都等於同一個常數：
${\displaystyle \cos ^{2}15^{\circ }+\cos ^{2}15^{\circ }-{\sqrt {3}}\sin 15^{\circ }\sin 15^{\circ }}$
${\displaystyle \cos ^{2}80^{\circ }+\cos ^{2}(-50^{\circ })-{\sqrt {3}}\sin 80^{\circ }\sin(-50^{\circ })}$
${\displaystyle \cos ^{2}170+\cos ^{2}(-140^{\circ })-{\sqrt {3}}\sin 170^{\circ }\sin(-140^{\circ })}$
(1) 求出這個常數。
(2) 將其推廣為一個三角恆等式，並給出證明。

## 參考資料

1. 人民教育出版社中學數學室. 第4章「三角函數」第2部分「兩角和與差的三角函數」第4.7節「二倍角的正弦、餘弦、正切」. 數學. 全日制普通高級中學教科書 (必修). 第1冊 (下) 1. 中國北京沙灘后街55號: 人民教育出版社. 2003: 42–47. ISBN 7-107-17105-4 （中文（中國大陸））.
2. 人民教育出版社中學數學室. 第4章「三角函數」第1部分「任意角的三角函數」第4.5節「正弦、餘弦的誘導公式」. 數學. 全日制普通高級中學教科書 (必修). 第1冊 (下) 1. 中國北京沙灘后街55號: 人民教育出版社. 2003: 28–33. ISBN 7-107-17105-4 （中文（中國大陸））.